Nyala is bungee jumping. The following function gives her elevation, in meters, $t$ seconds after jumping: $E(t)=\dfrac{80\text{sin}(0.6t)}{t}+15$ What is the instantaneous rate of change of Nyala's elevation after $5$ seconds? Choose 1 answer: Choose 1 answer: (Choice A) A $-9.96$ meters per second (Choice B) B $-9.96$ seconds per meter (Choice C) C $17.26$ meters per second (Choice D) D $17.26$ seconds per meter
Solution: Understanding the problem The function that represents the instantaneous rate of change of $E(t)$ is its derivative, $E'(t)$. Therefore, the instantaneous rate of change of the elevation after $5$ seconds is $E'(5)$. Let's find $E'(t)$ and evaluate it at $t=5$. Finding $E'(t)$ $E'(t)=\dfrac{48t\text{cos}(0.6t)-80\text{sin}(0.6t)}{t^2}$ Finding $E'(5)$ $\begin{aligned} E'({5})&=\dfrac{48({5})\text{cos}(0.6({5}))-80\text{sin}(0.6({5}))}{({5})^2} \\\\ &=\dfrac{240\text{cos}(3)-80\text{sin}(3)}{25} \\\\ &\approx -9.96 \end{aligned}$ Interpreting units $E(t)$ is Nyala's elevation in ${\text{meters}}$ after $t$ ${\text{seconds}}$. Therefore, we measure its rate of change in ${\text{meters}}$ per ${\text{seconds}}$. In conclusion, the instantaneous rate of change of Nyala's elevation after $5$ seconds is $-9.96$ meters per second. The rate of change is negative because Nyala is falling down at that time.